Problem: Simplify and expand the following expression: $ \dfrac{a + 8}{3a + 3}-\dfrac{2a + 9}{5a + 7} $
Explanation: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(3a + 3)(5a + 7)$ Multiply the first term by $\dfrac{5a + 7}{5a + 7}$ $ \begin{align*} \dfrac{a + 8}{3a + 3} \times \dfrac{5a + 7}{5a + 7} & = \dfrac{(a + 8)(5a + 7)}{(3a + 3)(5a + 7)} \\ & = \dfrac{5a^2 + 47a + 56}{(3a + 3)(5a + 7)}\end{align*} $ Multiply the second term by $\dfrac{3a + 3}{3a + 3}$ $ \begin{align*} \dfrac{2a + 9}{5a + 7} \times \dfrac{3a + 3}{3a + 3} & = \dfrac{(2a + 9)(3a + 3)}{(5a + 7)(3a + 3)} \\ & = \dfrac{6a^2 + 33a + 27}{(5a + 7)(3a + 3)}\end{align*} $ Now we have: $ = \dfrac{5a^2 + 47a + 56}{(3a + 3)(5a + 7)} - \dfrac{6a^2 + 33a + 27}{(5a + 7)(3a + 3)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{5a^2 + 47a + 56 - (6a^2 + 33a + 27)}{(3a + 3)(5a + 7)} $ $ = \dfrac{5a^2 + 47a + 56 - 6a^2 - 33a - 27}{(3a + 3)(5a + 7)} $ $ = \dfrac{-a^2 + 14a + 29}{(3a + 3)(5a + 7)}$ Expand the denominator: $ = \dfrac{-a^2 + 14a + 29}{15a^2 + 36a + 21}$